#352

Hard

auto_awesomeBinary search over the valid answer space

LeetCode Problem Workspace

Data Stream as Disjoint Intervals

The problem involves designing a class to summarize a data stream of non-negative integers as disjoint intervals using binary search.

binary search design ordered set

Problem Statement

Given a data stream of non-negative integers, design a system to summarize the numbers seen so far as a list of disjoint intervals. Each time a number is added, the system should return the updated set of intervals.

The task requires implementing a class called SummaryRanges. This class should handle two main operations: adding a number to the stream and retrieving the list of disjoint intervals. Efficient handling of these operations is key, particularly when dealing with large inputs.

Examples

Example 1

Input: See original problem statement.

Output: See original problem statement.

Input ["SummaryRanges", "addNum", "getIntervals", "addNum", "getIntervals", "addNum", "getIntervals", "addNum", "getIntervals", "addNum", "getIntervals"] [[], [1], [], [3], [], [7], [], [2], [], [6], []] Output [null, null, [[1, 1]], null, [[1, 1], [3, 3]], null, [[1, 1], [3, 3], [7, 7]], null, [[1, 3], [7, 7]], null, [[1, 3], [6, 7]]]

Explanation SummaryRanges summaryRanges = new SummaryRanges(); summaryRanges.addNum(1); // arr = [1] summaryRanges.getIntervals(); // return [[1, 1]] summaryRanges.addNum(3); // arr = [1, 3] summaryRanges.getIntervals(); // return [[1, 1], [3, 3]] summaryRanges.addNum(7); // arr = [1, 3, 7] summaryRanges.getIntervals(); // return [[1, 1], [3, 3], [7, 7]] summaryRanges.addNum(2); // arr = [1, 2, 3, 7] summaryRanges.getIntervals(); // return [[1, 3], [7, 7]] summaryRanges.addNum(6); // arr = [1, 2, 3, 6, 7] summaryRanges.getIntervals(); // return [[1, 3], [6, 7]]

Constraints

0 <= value <= 104
At most 3 * 104 calls will be made to addNum and getIntervals.
At most 102 calls will be made to getIntervals.

Solution Approach

Binary Search for Interval Insertion

To efficiently manage the intervals, use binary search to find where the new number fits. This allows quick insertion or merging with existing intervals, maintaining the disjoint property of the intervals.

Efficient Interval Management

Intervals should be stored in a sorted order, allowing quick retrieval and modification. When a number is added, check if it extends or merges existing intervals; otherwise, create a new interval.

Optimizing Space and Time Complexity

The algorithm must ensure O(log(N)) time complexity for the addNum operation and O(N) space complexity for storing intervals. This is achieved by efficiently managing the sorted list of intervals.

Complexity Analysis

Metric	Value
Time	O(log(N))
Space	O(N)

The time complexity of addNum is O(log(N)) due to the binary search used for insertion. Space complexity is O(N) because the intervals are stored in a list, which grows as more numbers are added to the stream.

What Interviewers Usually Probe

Assess whether the candidate understands how binary search can optimize interval management.
Look for an explanation of how to balance interval merging with insertion efficiency.
Evaluate the candidate's approach to managing sorted data and maintaining efficient performance.

Common Pitfalls or Variants

Common pitfalls

Failing to account for merging intervals properly when adding numbers.
Not using binary search effectively, leading to slower insertion times.
Misunderstanding the space complexity and overcomplicating the storage of intervals.

Follow-up variants

Handling negative numbers in the data stream.
Optimizing for concurrent access or thread safety in a multithreaded environment.
Modifying the problem to return intervals in a specific order or format.

FAQ

What is the main strategy for solving Data Stream as Disjoint Intervals?

The primary strategy is to use binary search to efficiently insert numbers into sorted intervals while managing disjoint ranges.

How does binary search improve performance in this problem?

Binary search helps find the correct position for inserting a number into the intervals in O(log(N)) time, making the solution scalable.

Can the interval merging step be optimized further?

Merging intervals efficiently is key. The binary search ensures the merging step remains efficient by minimizing unnecessary comparisons.

What is the time complexity of the addNum operation?

The time complexity of the addNum operation is O(log(N)) due to the use of binary search for insertion.

How does GhostInterview assist with the binary search pattern in this problem?

GhostInterview provides a focused explanation of binary search, helping you understand how to apply it effectively to the interval insertion and merging process.

terminal

Solution

Solution 1

#### Python3

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class SummaryRanges:
    def __init__(self):
        self.mp = SortedDict()

    def addNum(self, val: int) -> None:
        n = len(self.mp)
        ridx = self.mp.bisect_right(val)
        lidx = n if ridx == 0 else ridx - 1
        keys = self.mp.keys()
        values = self.mp.values()
        if (
            lidx != n
            and ridx != n
            and values[lidx][1] + 1 == val
            and values[ridx][0] - 1 == val
        ):
            self.mp[keys[lidx]][1] = self.mp[keys[ridx]][1]
            self.mp.pop(keys[ridx])
        elif lidx != n and val <= values[lidx][1] + 1:
            self.mp[keys[lidx]][1] = max(val, self.mp[keys[lidx]][1])
        elif ridx != n and val >= values[ridx][0] - 1:
            self.mp[keys[ridx]][0] = min(val, self.mp[keys[ridx]][0])
        else:
            self.mp[val] = [val, val]

    def getIntervals(self) -> List[List[int]]:
        return list(self.mp.values())


# # Your SummaryRanges object will be instantiated and called as such:
# # obj = SummaryRanges()
# # obj.addNum(val)
# # param_2 = obj.getIntervals()

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